LINDSTEDT'S METHOD 17

LINDSTEDT'S METHOD 17 As a check on the program and the method, we compare the results of LC with numerical integration on eq.(). Fig.3 shows the limit cycle of eq.() for c = 0.16 (i.e. ~ = 0.4 in the perturbation scheme, see eq.(3.» The curves in Fig.3 were obtained from the previous run (truncation of order 4), as well as from runs with truncation orders 6 and 8. In addition, Fig.3 displays numerical results generated by a Runge Kutta scheme. Note how, for this choice of c, greater accuracy is achieved by taking more terms in the series. Hopf Bifurcation As an application of the program LC, we investigate the Hopf bifurcation [36]. Whenever an equilibrium point of the focus type (i.e. having complex eigenvalues) changes its stability as a result of parameter changes in the differential equations, a limit cycle is generically born. In order to use Lindstedt's method to see how this occurs, we consider the most general version of eqs.(4) up to cubic terms, i.e. we expand f and g in Taylor series. We also include damping terms with coefficient M so that varying M through zero will change the stability of the equilibrium, producing the desired result: (5.1) x fxx fvv -y + c M x + c (~x + fxy x y + ~ y ) (fxxx 3 ~ ~!.ril 3) +c --6- x + xy + xy + 6 Y + (5.) y' <no< m, x + c M Y + c (~x + gxy x y + y)

18 LINDSfEIJf'S METHOD 1 x(t) o 8 t -1 Fig.3. Numerical (= N) and Lindstedt 0(8), 0(6) and 0(4) (= L8, L6 and L4 respectively) solutions for the limit cycle of eq.() for = 0.16 (Le. ' = 0.4).

LINDSTEDT'S METHOD 19 Here is the run: LC() ; The d.e. '5 are of the form x' = -y + emf, y' = x + ehg where f,g are of the form: quadratic + e*(linear + cubic) + e A *(quartic) +... Enter f: e*(fyyy*ya3/6+fxyy*x*ya/+fxxy*xa*y/+fxxx*xa3/6)+fyy*ya/+fxy*x*y +fxx*x A /+e*mu*x; f 3 3 fyyy y fxyy x y fxxy x y fxxx x fyy y fxx x e (------- + --------- + --------- + -------) + ------ + fxy x y + 6 6 Enter g: e*(gyyy*ya3/6+~*x*ya/+gxxy*xa*y/+gxxx*x~3/6)+gyy*ya/+gxy*x*y +gxx*x A /+e*mu*y; + e mu x g 3 3 gyyy y ~ x y gxxy x y gxxx x gyy y e (------- + --------- + --------- + -------) + ------ + gxy x y + e mu y 6 6 Enter truncation order: Done with step 1 of Choices for limit cycle amplitude: 1) 0 ) - 4 sqrt(- mu/(gyyy - gxy gyy + fyy gyy - gxx gxy + gxxy - fxx gxx gxx x + ------ + fxy fyy + fxyy + fxx fxy + fxxx» 3) 4 sqrt(- mu/(gyyy - gxy gyy + fyy gyy - gxx gxy + gxxy - fxx gxx Enter choice number 3; Done with step of + fxy fyy + fxyy + fxx fxy + fxxx» x = 4 cos(z) sqrt(- mu/(gyyy + (- gxy + fyy) gyy - gxx gxy + gxxy - fxx gxx + fxy fyy + fxyy + fxx fxy + fxxx» +.. y 4 sin(z) sqrt(- mu/(gyyy + (- gxy + fyy) gyy - gxx gxy + gxxy - fxx gxx + fxy fyy + fxyy + fxx fxy + fxxx» +..

0 LINDSTEDT'S METHOD w e ( gyy + (5 gxx - 5 fxy) gyy - 3 gxyy + gxy + (- fyy - 5 fxx) gxy - 3 gxxx + 5 gxx - fxy gxx + 3 fyyy + 5 fyy + 5 fxx fyy + fxy + 3 fxxy + fxx ) mu/(3 gyyy + (3 fyy - 3 gxy) gyy - 3 gxx gxy + 3 gxxy - 3 fxx gxx + 3 fxy fyy + 3 fxyy + 3 fxx fxy + 3 fxxx) + [VAX 8500 TIME = 331 SEC.] The significance of this computation lies in the requirement that the limit cycle amplitude be real. The limit cycle amplitude is given by the expression: (6) where limit cycle amplitude 4 (_~S)I/, (7) S = gyyy + gxxy + fxxx + fxyy + fyy gyy + fxy (fxx + fyy) - fxx gxx - gxy (gxx + gyy) In particular this requires that the sign of S be opposite to the sign of~. If S = 0 the computation is indecisive and offers no information. See Fig.4 for the two cases of subcritical and supercritical Hopf bifurcations, in which the stability of the limit cycle is respectively unstable and stable. Note that the form of the linear terms in eqs.(5) involves the symmetric appearance of ~ in both the x' and y' equations. A general system of the form (8) x' = a x + b y + y c x + d y + can be transformed into u v' = n u + ~ v +

LINDSTEDT'S METHOD 1 SUPERCRITICAL CASE, S < 0 y y x x ]1 < 0 A ]1 > 0 STABLE LIMIT CYCLE STABLE EQUILIBRIU~ UNSTABLE EQUILIBRIUM SUBCRITICAL CASE, S > 0 y y ---\--+--fd:h-4--x ---+-+-+++-++--- x ]1 < 0 A ]1 > 0 UNSTABLE LIMIT CYCLE STABLE EQUILIBRIUM UNSTABLE EQUILIBRIUM Fig.4. Supercritical and subcritical Hopf Bifurcations. A = amplitude of limit cycle.

LINDSTEDT'S METHOD via the linear transformation [36] (9) x = b u, y [ d-a] "" u"-flv where fl = [- (d~a)_ bcf/ and where Jl = a;d In this way we could apply the results (6),(7) to the system (30) x" + x + e c x' + e (a l x + a x x' + a 3 x' ) by writing it in the form: (31. I) x' - y (31.) y =x - e c y + e (a l x - a x y + a 3 y ) 3 3 + e (b x - b x y + b x y - by) + 1 3 4 and then using the transformation (9) to get it in the form (5). Alternately we can apply the program LC directly to the system (31): LC(); The d.e. 's are of the form x' =-y + emf, y' =x + e*g where f,g are of the form: quadratic + e*(linear + cubic) + e~*(quartic) +... Enter f: 0; f = 0 Enter g: A -c*y*e+(al*x~-a*x*y+a3*y~)+e*(bl*xa3-bmxa*y+b3*x*ya-b4*y 3); 3 3 g =e (- b4 y + b3 x y - b x y + bl x ) + a3 y - a x y - c e y + al x

Enter truncation order: : Done with step 1 of Choices for limit cycle amplitude: 1) 0 c ) - sqr t (- --------------------------) 3 b4 + b - a a3 - al a c 3) sqrt(- -------------------------) 3 b4 + b - a a3 - al a LINDSTEDT'S METHOD 3 Enter choice number 3; Done with step of c x = cos(z) sqrt(- -------------------------) +... 3 b4 + b - a a3 - al a c y sin(z) sqrt(- -------------------------) +... 3 b4 + b - a a3 - al a (3 b3 + 9 bl - 4 a3-10 al a3 - a - 10 al ) c e w 1 - ---------------------------------------------------- 18 b4 + 6 b - 6 a a3-6 al a [VAX 8500 TIME 74 SEC. ] From this run we can conclude that for small c eq.(30) possesses a limit cycle if the quantity (3) has the opposite sign to that of the damping c. For example. in the case of van der Pol's equation (). where c = -I, b = 1 and the other coefficients are zero. the quantity (3) becomes unity and the limit cycle amplitude is correctly evaluated as. Note also that the c contribution to the frequency w vanishes for van der Pol's equation. (Cf. eq.(15) where k 1 must be replaced by c for comparison with the foregoing run.) o and where c

4 LINDSTEDT'S METHOD Exercises 1. Find an equation of the form x" + x + e f(x.x ) 0 which exhibits a limit cycle with frequency w = 1 + k l e +. where k l # O. Note that the examples used in this Chapter. namely van der Pol's equation (). eq.(). and eq.(30). all have k l = O. Hint: Take a l =a =~ = 0 in eq.(30).. We have dealt exclusively with autonomous systems in this Chapter. Write a program to apply Lindstedt's method to nonautonomous equations of the form x" + x + e f(x.x.t) O. As an example. consider the forced damped Duffing equation. x + x + e 0 x + e a x 3 e..., cos wt Discussion: First we set T = wt to get d x dx 3 w -- + x + e 0 w - + e a x e..., cos T dt dt and then we expand x = XO(T) + xl(t) e +. We look for solutions with period v in T. i.e. with the same period as the forcing function.

LINDSTEUT'S METHOD 5 In order to eliminate secular terms in the xl equation we can now choose A O and B O ' (Note that the ki's are viewed as being given in this formulation.) Similarly the complementary solution for the Xi equation can be chosen as Xi = Ai sin T + B i cos T and an appropriate selection of Ai and comp B i will eliminate secular terms in the x i + 1 equation. Unfortunately the algebraic equations on A O and B O (or more generally on Ai and B i ) are usually too complicated to solve in closed form. Transformation to polar coordinates usually helps. E.g. in the case of Duffing's equation, A O and B O satisfy o which become in polar coordinates, which can be combined to give...

6 LINDSTEIJr'S MErHOn 3. Mathieu's equation. x" + (0 + 10 cos t) X 0 is both linear and nonautonomous. The problem here is to find the coefficients 0i in the expansion n 0.1.. such that the differential equation exhibits periodic solutions. This expansion represents a transition curve in the 0-10 parameter plane (hence the phrase parametric excitation) which separates regions of bounded behavior from regions of unbounded behavior. See e.g. [44]. Write a program to accomplish this task. Note that since the equation is linear. no frequency-amplitude relation is expected and no stretch T = ~ t is needed. so that regular perturbations will work on this problem. After expanding x = X o + Xl 10 +. separately choose X n o = sin t and then X o = cos ~t. as each gives a distinct transition curve. Find the 0i by elimination of secular terms. See [35]. 4. Apply the transformation (9) to eqs.(3i) to get them in the form (5). Then show that the results (6) and (7) agree with results of the last run in the Chapter on eqs.(3i) directly.

CHAPTER CENTER MANIFOLDS Introduction Center manifold theory [6] is a method which uses power series expansions in the neighborhood of an equilibrium point in order to reduce the dimension of a system of ordinary differential equations. The method involves restricting attention to an invariant subspace (the center manifold) which contains all of the essential behavior of the system in the neighborhood of the equilibrium point as t ~ 00. This method is applicable to systems which, when linearized about an equilibrium point, have some eigenvalues which have zero real part, and others which have negative real part. We assume that no eigenvalues have positive real part, since in such a case the center manifold will not be attractive as ~oo Under such assumptions the components of the solution of the linearized equations which correspond to those eigenvalues with negative real part will decay as t ~ 00, and hence the motion of the linearized system will asymptotically approach the space 8 spanned by the eigenvectors corresponding 1 to those eigenvalues with zero real part. The center manifold theorem [6] assures us that this picture (which is so far based on the linearized equations) extends to the full nonlinear equations. as follows:

8 CENTER MANIFOLfS There exists a (generally curved) subspace S (the center manifold) which is tangent to the (flat) subspace SI at the equilibrium point. and which is invariant under the flow generated by the nonlinear equations. All solutions which start sufficiently close to the equilibrium point will tend asymptotically to the center manifold. In particular. the theorem states that the stability of the equilibrium point in the full nonlinear equations is the same as its stability when restricted to the flow on the center manifold ([6J. p.4). Moreover. any additional equilibrium points or limit cycles which occur in a neighborhood of the given equilibrium point on the center manifold are guaranteed to exist in the full nonlinear equations ([6J. p.9). Thus center manifold theory allows us to eliminate algebraic complications which are due to unessential behavior and to focus on the motion in the center manifold which contains the critical information concerning the system's local stability and bifurcation of steady state behavior. In doing so we omit reference to how the system gets onto the center manifold. but rather we are assured of the asymptotic stability of the center manifold. Since the eigenvalues associated with the linearized flow in the center manifold have zero real part. the study of the motion in the center manifold must still be accomplished and may not be an easy task. In this Chapter we will be concerned with the question of how to reduce a system to its center manifold and we will leave for later Chapters treatment of the flow on the center manifold. The method will be illustrated by reference to the following example [3IJ of a system of three differential equations: (1) x y () y' - x - x z (3) z - z + a x

CENTER MANIFOLDS 9 Here we have a feedback control system in which the spring constant in the simple harmonic oscillator of eqs.(1).() is given by 1 + z. The control variable z is governed by eq.(3) and involves the feedback term a x. We wish to know the stability of the equilibrium at the origin. Although the system (1)-(3) is 3-dimensional. we can use center manifold theory to reduce the stability question to the study of a -dimensional system. Linearizing about the origin. we see that the x-y plane is associated with a pair of pure imaginary eigenvalues, while the z axis corresponds to an eigenvalue of -1. Thus the center manifold is a -dimensional surface which is tangent to the x-y plane at the origin. In order to obtain an approximate expression for the center manifold, we write z as a function of x and y, and expand in a power series. Since the center manifold is tangent to the x-y plane, we begin the power series with quadratic terms: (4) z a,o x + a 1,1 x y + a O, y + The requirement that the center manifold be invariant is satisfied by substituting (4) into (3): (5) ( a. 0 x + a 1,1 y) x' + ( a O, y + al,l x) y + - (a,o x + a l. 1 x y + a O. y ) + a x + Note that (5) involves the derivatives x and y'. Substitution of expressions for these from (1) and () gives:

30 CENTER MANIFOLDS (6) ( a 0 x + a l. l y) y + ( a O. y + a l. l x) (- x - x z) + - (a. 0 x + a l. l x y + a O. y ) + a x + This last step once again introduced z into the computation (since y depended on z). So we again eliminate z by substituting (4) into (6). Collecting terms and neglecting cubic and higher order terms. this gives: (7) + (a l. l + a O. ) y + = 0 Equating the coefficients of xiyj equal to zero. and solving the three resulting equations for the a.. s. we obtain: 1. J (8) 3 a 0 = g a - g a. a O = g a center manifold: Substitution in (4) gives the following first approximation to the (9) 3 z=gax -gaxy+gay + We may now substitute (9) into eqs.(1).() in order to obtain approximate equations for the flow on the center manifold: (10) x = y (11 ) y' 3 3 -x-gax +gax y-gaxy +

CENTER MANIFOLDS 31 Although approximate. eqs.(lo).(ll) contain the answer to the question of the stability of the origin in the original system (1)-(3). The point of center manifold theory is that eqs.(lo).(ll) are easier to analyze than eqs.(1)-(3). We will return to the analysis of eqs.(lo).(ll) in the next Chapter on normal forms. Computer Algebra The calculation of center manifolds involves the manipulation of truncated power series and is readily performed using computer algebra. We present a MACSYMA program to accomplish such computations. The program listing follows a sample run on the preceding example: CM() ; ENTER NO. OF EQS. 3; ENTER DIMENSION OF CENTER MANIFOLD ; THE D.E. 'S MUST BE ARRANGED SO THAT THE FIRST EQS. REPRESENT THE CENTER MANIFOLD. I.E. ALL ASSOCIATED EIGENVALUES ARE ZERO OR HAVE ZERO REAL PARTS. ENTER SYMBOL FOR VARIABLE NO. X; ENTER SYMBOL FOR VARIABLE NO. Y; ENTER SYMBOL FOR VARIABLE NO. 3 Z: ENTER ORDER OF TRUNCATION : ENTER RHS OF EQ. DxIDT= Y: ENTER RHS OF EQ. D y IDT -X-X*Z: ENTER RHS OF EQ. 3 D z IDT = -Z+ALPHA*X A :

3 CENTER MANIFOLDS dx -- = Y dt dy -- = - x z - x dt dz -- =alpha x - z dt CENTER MANIFOLD: alpha y alpha x y 3 alpha x [z =---------- - ----------- + ----------] 5 5 5 FLOW ON THE C.M.: dx dy alpha y alpha x y 3 alpha x [-- = y. -- = - x (---------- - ----------- + ----------) - x] dt dt 5 5 5 [VAX 8500 TIME = 6 SEC.] Here is the MACSYMA listing for the program CM: CM() :=( 1* INPUr PROBLEM *1 N:READ("ENTER NO. OF EQS."). M: READ( "ENTER DIMENSION OF CENTER MANIFOLD"). PRINT("THE D.E. 'S MUST BE ARRANGED SO THAT THE FIRST".M. "EQS."). l'rint("represent THE CENTER MANIFOLD. I.E. ALL ASSOCIATED"). PRINT("EIGENVALUES ARE ZERO OR HAVE ZERO REAL PARTS."). FOR I: 1 THRU N 00 X[I]:READ("ENTER SYMBOL FOR VARIABLE NO... I). L:READ("ENTER ORDER OF TRUNCATION"). FOR 1:1 THRU N DO ( PRINT("ENTER RHS OF EQ... I). PRINT("D".X[I]."/DT =").

CENTER MANIFOLDS 33 G[ IJ: READ() ). 1* SET UP D.E. 'S *1 FOR I: 1 THRU N DO DEPENDS(X[I].T). FOR I: 1 THRU N DO (EQ[I]:DIFF(X[I].T)=G[I]. PRINT(EQ[I]». 1* FORM POWER SERIES *1 SUB:MAKELIST(K[I].I.l.M). VAR:PRODUCT(X[I]AK[I].I.l.M). UNK:[]. FOR P:M+l THRU N DO{ TEMP:A[P.SUB]*VAR. FOR I: 1 THRU MDO TEMP:SUM(EV(TEMP.K[I]=J).J.O.L). TEMP:TAYLOR(TEMP.MAKELIST(X[I].I.l.M).O.L). 1* REMOVE ronstant AND LINEAR TERMS *1 TEMP3:TEMP-PART(TEMP,l)-PART(TEMP.). SOLN[P]:EXPAND(TEMP3). 1* PREPARE LIST OF UNKNOWNS *1 SETXTOl:MAKELIST(X[I]:l.I.l.M). 1* TURN SUM INTO A LIST *1 UNKN[P]:SUBST(..[...+... EV(TEMP3.SETXTOl». UNK:APPEND(UNK.UNKN[P]». SOL:MAKELIST(X[P]:SOLN[P].P.M+l.N). 1* SUBSTITUTE INTO D. E..S *1 CMDE:MAKELIST(EQ[I].I.l.M). REST:MAKELIST(LHS(EQ[I])-RHS(EQ[I]).I.M+l.N). TEMP4:EV(REST.SOL. DIFF),

34 CENTER MANIFOLDS TEMP5: EV (TEMP4, CMDE,DIFF), TEMP6:EV(TEMP5,SOL). TEMP7:TAYLOR(TEMP6,MAKELIST(X[I].I.1,M),O,L), 1* COLLECT TERMS *1 COUNTER: 1, 1* MAKE LIST OF TERMS *1 TERMS:SUBST("[", "+",SOLN[NJ). TERMS:EV(TERMS.A[DUMMY,SUB]:=l). FOR I: 1 THRU N-M DO ( EXP[I]:EXPAND(PART(TEMP7.I». FOR J: 1 THRU LENGTII(TERMS) DO( COND[COUNTER]:RATCOEF(EXP[I].PART(TERMS.J». COUNTER: COUNTER+ 1) ) CONDS:MAKELIST(COND[I],I,l.COUNTER-1). 1* SOLVE FOR CENTER MANIFOLD *1 ACOEFFS:SOLVE(CONDS. UNK). CENTERMANIFOLD: EV(SOL,ACOEFFS) PRINT("CENTER MANIFOLD: "). PRINT(CENTERMANIFOLD) 1* GET FLOW ON CM *1 CMDE: EV (CMDE, CENTERMANI FOLD) PRINT("FLOW ON THE C.M. :"). PRINT(CMDE»$

CENTER MANIFOLDS 35 Systems with Damping Since the center manifold must contain a linearized system with eigenvalues equal to zero or having zero real part. it would appear that the method is inapplicable to the systems which have damping. Carr [6] has shown. however. that such cases can be handled by embedding the given system in a larger system which contains the damping as an additional dependent variable. For example. consider the previous example when damping is added: (1) x' ~ x + y (13) y - x + ~ y - x z (14) z Now to these equations we add the dummy equation: (15) o The system (1)-(15) is now a 4-dimensional system with a 3-dimensional center manifold. The damping terms ~ x and ~ yare now nonlinear (quadratic) terms in the neighborhood of the origin x = y = z = ~ = O. Naturally the results of the method are only valid for small values of ~. In order to illustrate the procedure. we apply the program CM to this example:

36 CENTER MANIFOLDS CM() ; ENTER NO. OF EQS. 4; ENTER DIMENSION OF CENTER MANIFOLD 3; THE D.E. 'S MUST BE ARRANGED SO THAT THE FIRST 3 EQS. REPRESENT THE CENTER MANIFOLD. I.E. ALL ASSOCIATED EIGENVALUES ARE ZERO OR HAVE ZERO REAL PARTS. ENTER SYMBOL FOR VARIABLE NO. MU; ENTER SYMBOL FOR VARIABLE NO. X; ENTER SYMBOL FOR VARIABLE NO.3 Y; ENTER SYMBOL FOR VARIABLE NO.4 Z; ENTER ORDER OF TRUNCATION 3; ENTER RHS OF EQ. D mu lot = 0; ENTER RHS OF EQ. D x lot = MU*X+Y; ENTER RHS OF EQ. 3 D y lot = MU*Y-X-X*Z; ENTER RHS OF EQ. 4 D z lot = -Z+ALPHA*X A ; dmu --- = 0 dt dx --=y+mux dt dy -- = - x z + mu y - x dt dz -- = alpha x - z dt

CENTER MANIFOLDS 37 CENTER MANIFOLD: 8 alpha mu y alpha y 8 alpha mu x y alpha x y [z -------------- + ---------- + -------------- - ----------- 5 5 5 5 FLOW ON THE C.M.: alpha mu x 3 alpha x - -------------- + ----------] 5 5 dmu dx dy 8 alpha mu y alpha y [--- = O. = y + mu x, - x (- -------------- + ---------- dt dt dt 5 5 8 alpha mu x y alpha x y alpha mu x 3 alpha x + -------------- - ----------- - -------------- + ----------) + mu y - x] 5 5 5 5 [VAX 8500 TIME 51 SEC.] That is. for small values of ~. the flow near the origin on the center manifold of the system (1)-(15) is given by: (16) x = y + ~ x ( 17) y Here the origin is unstable for ~ ) 0 and stable for ~ < O. Thus there is a Hopf bifurcation at ~ = O. Let us use the Lindstedt method program LC of Chapter 1 to approximate the size of the limit cycle as a function of the parameters a and~. In order to put the problem in a form suitable for treatment by LC. we scale x. y and ~ as follows: (18) X to x. Y - to y. M whereupon eqs. (16). (17) become:

38 CENTER MANIFOLDS (19) X'. [3 ] (0) Y = X + M c Y + a c X 5" X + 5" X Y + 5" Y + We continue with the application of the program LC to eqs.(19).(0): LC(); The d.e. '5 are of the form x' -y + e*f. y = x + e*g where f.g are of the form: quadratic + e*(linear + cubic) + e A *(quartic) +... f = e m x Enter g: M*E*Y+ALPHA*E*X*(3*X A /S+*X*Y/S+*Y A /S); g = alpha e x y (---- + S x y 3 x + ----) S S + e m y Enter truncation order: ; Done with step 1 of Choices for limit cycle amplitude: 1) 0 m ) - sqrt(s) sqrt(- -----) alpha m 3) sqrt(s) sqrt(- -----) alpha Enter choice number 3; Done with step of

CENTER MANIFOLDS 39 m x sqrt(- -----) sqrt(5) cos(z) +... alpha m y sin(z) sqrt(- -----) sqrt(5) +... alpha w 1-11 e m [VAX 8500 TIME 1 SEC.] That is, to lowest order approximation, the limit cycle amplitude is calculated as (cf. eq.(18): 5M] 1/ (1) [- - or a ( x + y )1/ Eq.(1) asserts that for small ~ the limit cycle exists only if ~ and a have opposite signs. As a check on this computation, we present in Fig.5 the results of a numerical integration of the differential equations (1)-(14) for parameter values ~ = 0.01, a = -0.01. As can be seen from the Figure, the foregoing result based on center manifold and Lindstedt methods is in good agreement with the numerical computation.

40 CENTER MANIFOLDS 5 y -5 5 x -5 Fig.~. Numerical (= N) and center manifold/ Lindstedt 0() (= em) solutions for the limit cycle of eqs.(1)-(14) for ~ = 0.01 and a = -0.01, displayed in the x-y plane.

CENTER MANIFOLDS 41 Eigencoordinates The program CM requires that the coordinates on the center manifold be linearly uncoupled from the other coordinates of the problem. In the case that the original system is not in this form. some linear algebra needs to be done. We shall give an example of this complication next. as we consider a bifurcation in the well-known Lorenz equations [14]: () x a (y - x) (3) y px-y-xz (4) z' - /3 z + x y We shall be interested in the behavior of this system in the neighborhood of the equilibrium at the origin. Linearizing about the origin. we obtain (5) a -1 o J] [~ ] We use the EIGEN package in MACSYMA to compute the eigenvalues of the linearized system: (Here and in what follows we use the symbols s.r.b and e in MACSYMA to stand for a.p./3 and c respectively. We refer to the matrix of the linearized system (5) as A.) A:MATRIX([-S.S.O].[R.-1.0].[O.O.-B]); [ - s s 0 ] [ ] [ r - 1 0 ] [ ] [ 0 0 - b ]

4 CENTER MANIFOLDS EIGENVALUES(A) ; sqrt(s + (4 r - ) s + 1) + s + 1 [[- ---------------------------------- sqrt(s + (4 r - ) s + 1) - s - 1 - b]. [1, 1. 1]] Here the first list of three eigenvalues is followed by a second list of their respective multiplicities. Note that at p = lone of the eigenvalues is zero for all a and~. Thus at p = 1 we have a one-dimensional center manifold. We propose to use our program CM to study the bifurcations of equilibria in the neighborhood of the origin. Before we can proceed. however. we must uncouple the eigencoordinate corresponding to the center manifold from the other eigencoordinates. We return to our previous MACSYMA session and compute the eigenvectors of the linearized system when p = 1: ElGENVECfORS(EV(A.R:l»; [[[- s - 1. - b. 0]. [1. 1. 1]]. [1. - - 0]. [0. O. 1]. [1. 1. 0]] s Here the first list contains the eigenvalues. the second their respective multiplicities. and the last three lists are their respective eigenvectors. In order to diagonalize the linearized system. we transform variables from x.y.z to u.v.w via a matrix whose columns are the preceding eigenvectors: (6) Now we need to substitute (6) into ()-(4). We will use our own MACSYMA utility program. TRANSFORM. to accomplish this. TRANSFORM performs an

CENfER MANIFOLDS 43 arbitrary coordinate transformation (not necessarily linear) and will be used again in later Chapters. We follow the MACSYMA run with a listing of TRANSFORM: TRANSFORM() ; Enter number of equations 3; Enter symbol for original variable X Enter symbol for original variable Y; Enter symbol for original variable 3 Z; Enter symbol for transformed variable U; Enter symbol for transformed variable V; Enter symbol for transformed variable 3 W; The RHS's of the d.e. 's are functions of the original variables: Enter RHS of x d.e. d x Idt = S*(Y-X) ; d x Idt = s (y - x) Enter RHS of y d.e. d y Idt = R*X-Y-X*Z; d y Idt = - x z - y + r x Enter RHS of z d.e. d z Idt = -B*Z+X*Y; d z Idt = x y - b z The transformation is entered next: Enter x as a function of the new variables x = U+V; x = v + u Enter y as a function of the new variables y = U-V/S; v y u - s

44 CENTER MANIFOLDS Enter z as a function of the new variables z = W; z =w du 5 (u W + (1 - r) u) + 5 V (w - r + 1) [[-- ------------------------------------- dt 5 + 1 dv 5 «r - 1) u - u w) + v (5 (- W + r + 1) + 5 + 1) dt 5 + I dw 5 (b w - u ) + v + (u - 5 u) V -------------------------------]] dt 5 [VAX 8500 TIME = SEC.] Here is the listing of TRANSFORM: TRANSFORM():=( 1* input data *1 n:read("enter number of equations"), for i:1 thru n do x[i]:read("enter symbol for original variable",i), for i:1 thru n do y[i]:read("enter symbol for transformed variable",i), print("the RHS's of the d.e.'s are functions of the original variables:"), for i:1 thru n do ( print("enter RHS of",x[i],"d.e."), print("d",x[i],"/dt ="), f[i]: read(), print("d",x[i],"/dt =",f[i]), print("the transformation is entered next:"), for i:1 thru n do ( print("enter",x[i],"as a function of the new variables"),

CENTER MANIFOLDS 45 print(x[i],":"), g[i J: read(), print(x[i]. ":".g[i]». 1* do it *1 for i:l thru n do depends([x[i],y[i]],t). for i:l thru n do eq[i]:diff(x[i],t):f[i]. trans:makelist(x[i]:g[i].i,i,n), for i:l thru n do treq[i]:ev(eq[i],trans,diff). treqs:makelist(treq[i].i.l,n), derivs:makelist(diff(y[i],t),i,l.n), neweqs:solve(treqs.derivs»$ In order to observe the bifurcation of equilibria in the center manifold as p passes through unity, we set (7) p 1 + /0 and we embed the system in a 4-dimensional phase space with /0' : O. The following MACSYMA command substitutes (7) into the list NEWEQS which contains the results of TRANSFORM. are stored in an array RHS, The right hand sides of the transformed equations to be conveniently passed on to the program CM: FOR 1:1 THRU 3 DO RHS[I]:EV(PART(NEWEQS.l.I.),R:l+E); CM() ; ENTER NO. OF EQS. 4; ENTER DIMENSION OF CENTER MANIFOLD ; THE D.E. 'S MUST BE ARRANGED SO THAT THE FIRST EQS. REPRESENT THE CENTER MAN IFOLD. I. E. ALL ASSOCIATED EIGENVALUES ARE ZERO OR HAVE ZERO REAL PARTS.

46 CENTER MANIFOLDS ENTER SYMBOL FOR VARIABLE NO. E; ENTER SYMBOL FOR VARIABLE NO. U; ENTER SYMBOL FOR VARIABLE NO.3 V; ENTER SYMBOL FOR VARIABLE NO. 4 W; ENTER ORDER OF TRUNCATION ; ENTER RHS OF EQ. D e IDT = 0; ENTER RHS OF EQ. D u IDT = RHS[1]; ENTER RHS OF EQ. 3 D v IDT = RHS[] ; ENTER RHS OF EQ. 4 D w IDT = RHS[3]; de -- =0 dt du 5 (u W - e u) + 5 V (w - e) dt 5 + dv 5 (e u - u w) + v (5 (- W + e + ) + 5 + 1) dt 5 + 1 dw 5 (b w - u ) + v + (u - 5 u) V dt 5 CENTER MANIFOLD: e 5 U U [v =- ------------, w =--] b 5 + 5 + 1

CENTER MANIFOLDS 47 FLOW ON THE C.M. : de [-- dt O. du dt 3 u s(---eu) b u e s u (-- - e) b s + s + 1 ------------------------------] s + 1 [VAX 8500 TIME 13 SEC.] The resulting expression from CM. stored in the variable CMDE. can be cleaned up by using the MACSYMA function FACTOR: FACTOR(CMDE) ; de [-- dt O. du dt s (s - e s + s + 1) u (u - b e) -----------------------------------] 3 b (s + 1) The last equation gives an approximation for the flow on the center manifold. There are 1 or 3 equilibria. depending on the sign of the product (3 10: (8) u o and u = + «(3 10)1/ Thus there is a pitchfork bifurcation at p = 1. see Fig.6. In order to check the center manifold computation. we calculated the center manifold numerically. It is straightforward to do this since any initial condition close to the origin produces a motion which asymptotically approaches the center manifold. In Fig.7 we plot in the x-y plane the numerical results for Lorenz's original parameter values of a = 10, (3 = 8/3, and for p 0.9. Fig.7 also shows the results of the preceding analysis. obtained as follows: We invert the change of coordinates (6) and transform the center manifold equations:

48 CENTER MANIFOLDS u Fig.6. Pitchfork bifurcation of equilibria on the center manifold at the origin in the Lorenz equations, see eq.(8). 1 y em x -1 Fig.7. Numerical (= N) and center manifold (= CM) solutions for the center manifold at the origin in the Lorenz equations for a = 10, S = 8/3, p = 0.9, E = -0.1, displayed in x-y plane.

CENrER MANIFOLDS 49 (9) v to 0 u - (1+0) u w = 73 giving (30) y (0+1) + to (0+1) - to 0 x _ (0+1)4 z - f3 [(0+1) - to x 0] For 0 = 10, f3 8/3 and to -0.1 this projects onto the x-y plane as the straight line (31) y 0.99... x Exercise 1. This problem concerns the behavior of van der Pol's equation at infinity [19]. If in van der Pol's equation, (PI) w we set [7]: (P) v = w / w and z 1 / w and reparameterize time with dt = w dt, we obtain: (P3.l) v (P3.) z 3 - v z where primes now represent derivatives with respect to T. The equilibrium at infinity, v = z = 0, has eigenvalues zero and -to. There is a center manifold tangent to the z-axis. Use the program CM to obtain an approximation f or it to O(z 1). Note that unlike the treatment of van der Pol's equation by Lindstedt's method, we do not assume to is small here.